Below is a question from an AP Calculus exam. The answer key say choice C is the correct answer, so that implies that $$\lim_{x\to1} (f(x)g(x+1))$$ does exist. It seems to me that all the choices are true, and there is no correct answer.
Question 1) If $\lim_{x\to1} (f(x)g(x+1))$ does exist then what is its value?
Question 2) Since it does exist, does that imply that $\lim_{x\to1} g(x+1)$ also exist?
Question 3) Isn't it true that:
$$\lim_{x\to1} g(x+1) = \lim_{x\to2} g(x) $$
and it is established that $\lim_{x\to2} g(x) $ doesn't exist in choice (b)?
This is my reasoning:
$$\lim_{x\to1} (f(x)g(x+1))$$
$$[\lim_{x\to1}f(x)] \times [\lim_{x\to1} g(x+1)]$$
$$[\lim_{x\to1}f(x)] \times [\lim_{x\to2} g(x) ]$$
$$[0] \times [DNE]$$
$$DNE$$
So there is either something I don't understand about limits, or the question is wrong. I want to say the question is wrong, but I'm not 100% confident.
Please Help.
Best Answer
Note that $|g(x)| \leq 1$
$$0 \leq |f(x)g(x+1) | \leq |f(x)|$$
Now we can apply squeeze theorem and show that
$$0 \leq \lim_{x \to 1} |f(x)g(x+1)| \leq \lim_{x \to 1} |f(x)| = 0 $$
We do not require $\lim_{x \to 1} g(x+1) $ to exists.
An extreme example would be $h(x) =0$ and $g(x)$ is some bounded function. Regardless of what is $g(x)$ exactly, we always have $h(x) g(x) = 0$.